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AlladinOne [14]
3 years ago
7

Y=74(1.01)^8 is it increasing or decreasing

Mathematics
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

Increasing

Step-by-step explanation:

The computation is given below:

Given that

y = 74 × (1.01)^8

= 74 × 1.082856706

= 80.1314

As it can be seen that the initial amount is 74 but after solving the given equation the value is increased as it shows 80.1314

Therefore it is increasing

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Is -451 in the sequence below?<br> 15,7,-1,-9,,-17..
Andrews [41]

Answer:

No, the closest to -451 in the sequence is -429

Step-by-step explanation:

5 0
3 years ago
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8. A rectangle's sides are measured to be 6.2 cm and 9.3 cm. What is the
Zepler [3.9K]
A= 9.3 * 6.2
A = 57.66cm2
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3 years ago
2 ÷ 48.9 plss need help
Mkey [24]

Answer:

0.04

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3 years ago
HELP ASAP SHOW YOUR WORK AND EXPLAIN HOW YOU GOT YOUR ANSWER-NO LINKS OR JOKING AROUND
Alexxx [7]

Step-by-step explanation:

1) -5x<35

we divide both sides by 5

-5x/5<35/5

-x<7

we divide both sides by-1 to remove the negative sign

-x/-1<7/-1

x<7/-1

2) 2x>-42

we divide both sides by 2

2x/2>-42/2

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3) x/3≤-7

we multiply by 3 and we cancel 3

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x≤-21

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8 0
3 years ago
Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card an
ziro4ka [17]

Answer:

1) is not possible

2) P(A∪B) = 0.7

3) 1- P(A∪B) =0.3

4) a) C=A∩B' and P(C)= 0.3

b)  P(D)= 0.4

Step-by-step explanation:

1) since the intersection of 2 events cannot be bigger than the smaller event then is not possible that P(A∩B)=0.5 since P(B)=0.4  . Thus the maximum possible value of P(A∩B) is 0.4

2) denoting A= getting Visa card , B= getting MasterCard the probability of getting one of the types of cards is given by

P(A∪B)= P(A)+P(B) - P(A∩B) = 0.6+0.4-0.3 = 0.7

P(A∪B) = 0.7

3) the probability that a student has neither type of card is 1- P(A∪B) = 1-0.7 = 0.3

4) the event C that the selected student has a visa card but not a MasterCard is given by  C=A∩B'  , where B' is the complement of B. Then

P(C)= P(A∩B') = P(A) - P(A∩B) = 0.6 - 0.3 = 0.3

the probability for the event D=a student has exactly one of the cards is

P(D)= P(A∩B') + P(A'∩B) = P(A∪B) - P(A∩B) = 0.7 - 0.3 = 0.4

3 0
3 years ago
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